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2 years ago

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A doctor's office schedules appointments in 30-minute time slots. All appointments are 30 minutes long. The office is open 8 hou

rs per day, 5 days a week. How many appointments can they schedule, in total, for Thursday and Friday?

1 answer:

horrorfan [7]2 years ago

3 0

8 hours per day.

16 total hours.

32 30 minute sessions!

Thus 32 appointments in total.

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Evaluate the following integral (Calculus 2) Please show step by step explanation!

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$4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x$

Rewrite 9 as 3²:

$\implies \displaystyle \int \dfrac{4}{x\sqrt{3^2+(\ln(x))^2}}\:\:\text{d}x$

<u>Integration by substitution</u>

$\boxed{\textsf{For }\sqrt{a^2+x^2} \textsf{ use the substitution }x=a \tan\theta}$

$\textsf{Let } \ln x=3 \tan \theta$

$\begin{aligned}\implies \sqrt{3^2+(\ln x)^2} & =\sqrt{3^2+(3 \tan\theta)^2}\\ & = \sqrt{9+9\tan^2 \theta}\\ & = \sqrt{9(1+\tan^2 \theta)}\\ & = \sqrt{9\sec^2 \theta}\\ & = 3 \sec\theta\end{aligned}$

Find the derivative of ln x and rewrite it so that dx is on its own:

$\implies \ln x=3 \tan \theta$

$\implies \dfrac{1}{x}\dfrac{\text{d}x}{\text{d}\theta}=3 \sec^2\theta$

$\implies \text{d}x=3x \sec^2\theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned} \implies \displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x & = \int \dfrac{4}{3x \sec \theta} \cdot 3x \sec^2\theta\:\:\text{d}\theta\\\\ & = \int 4 \sec \theta \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 4 \int \sec \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{7 cm}\underline{Integrating $\sec kx$}\\\\$\displaystyle \int \sec kx\:\text{d}x=\dfrac{1}{k} \ln \left| \sec kx + \tan kx \right|\:\:(+\text{C})$\end{minipage}}$

$\implies 4\ln \left| \sec \theta + \tan \theta \right|+\text{C}$

$\textsf{Substitute back in } \tan\theta=\dfrac{1}{3}\ln x :$

$\implies 4\ln \left| \sec \theta + \dfrac{1}{3}\ln x \right|+\text{C}$

$\textsf{Substitute back in } \sec\theta=\dfrac{1}{3}\sqrt{9+(\ln x)^2}:$

$\implies 4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}$

Learn more about integration by trigonometric substitution here:

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2 years ago